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Can anyone prove this question ?

Let $f$:$\mathbb{R}$$→$$\mathbb{C}$ be continuous function such that $f$$(0)$ $=$ $0$ and that $f'$ be a piecewise continuous function and absolutely integrable on $\mathbb{R}$.

Prove that

$L[f'](s)$ $=$ $sL[f](s)$

where $L[f]$ represents the Laplace transform of f

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Did you try integration by parts? – mrf May 6 '13 at 21:40
what would be my u and v values, if I applied the integration by parts method – S F May 6 '13 at 21:41
Why don't you begin by writing down the definition of the Laplace transform? – wj32 May 6 '13 at 21:41
$f(s)= $ $L(ft)(s)$ $$ \int^inf_0 dx $ – S F May 6 '13 at 21:44
And what does the LHS of the identity you're trying to prove look like? – wj32 May 6 '13 at 21:47
up vote 2 down vote accepted

$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$

Integrating by parts we have,

$L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$


If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$


Since $f(0)=0$, this reduces to


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thanks, do you know of any online resource that deals with this type of question ? – S F May 6 '13 at 21:54
I think Wiki is sufficient when studying basic Laplace theorem proofs. In the case of this question, wiki goes:… – Maazul May 6 '13 at 22:02

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